VD6:
2NaOH+FeCl2->Fe(OH)2+2NaCl
0,4.........0,15.........0,15........0,3
Lập tỉ lệ. 0,4/2:0,15/1=0,2>0,15. NaOH dư
mNaCl=0,3.58,5=17,55g
nNaOH dư=(0,4-0,3).40=4g
mdd sau pu=200+150=350g
C% NaCl...
C% NaOH dư ...
4Fe(OH)2+O2+H20->2Fe2O3+5H2O
0,15............................0,075
mFe2O3=0,075.160=12g